binomial expansion: What is the 4th term of (2x-3y) ^10

- C (10,3)  * (2x)⁷ (3y)³   =

- 120 * 128x7 * 27y3  =

-414720 x⁷ y³

binomial expansion: What is the 4th term of (2x-3y)^10

$\begin{array}{rrcr} & (2x-3y)^{10} = \\\\ 1\text{st term:} & \dbinom{10}{0} \cdot (2x)^{10} \cdot (-3y)^0 \\ 2\text{nd term:} & +\dbinom{10}{1} \cdot (2x)^{9} \cdot (-3y)^1 \\ 3\text{rd term:} & +\dbinom{10}{2} \cdot (2x)^{8} \cdot (-3y)^2 \\ 4\text{th term:} & +\dbinom{10}{3} \cdot (2x)^{7} \cdot (-3y)^3 &=& \dfrac{10}{3} \cdot \dfrac{9}{2} \cdot \dfrac{8}{1} \cdot 2^7\cdot x^7 \cdot (-3)^3 \cdot y^3\\ &&=& \dfrac{10}{3} \cdot \dfrac{9}{2} \cdot \dfrac{8}{1} \cdot 2^7\cdot x^7 \cdot (-27) \cdot y^3\\ &&=& \dfrac{10}{3} \cdot \dfrac{9}{2} \cdot \dfrac{8}{1} \cdot 128\cdot x^7 \cdot (-27) \cdot y^3\\ &&=& 10 \cdot 3 \cdot 4 \cdot 128\cdot x^7 \cdot (-27) \cdot y^3\\ &&=& -27 \cdot 10 \cdot 3 \cdot 4 \cdot 128\cdot x^7 \cdot y^3\\ &&=& -414720\cdot x^7 \cdot y^3\\ \dots & \dots \end{array}$